# Analytical Chemistry

How many milliliters of 0.202 M KOH should be added to 500.0 mL of 0.0233 M
tartaric acid (C4H6O6; FW 150.087) to adjust the pH to 2.75? Ka1 = 1.0 x 10-13 and Ka2 = 4.6 x10-5.

For this one, the Ka1 seems sort of off. The pka would be 13 which doesn't make sense.

1. 👍 0
2. 👎 0
3. 👁 46
1. Tartaric acid is a dibasic acid (two carboxyl COOH groups). You want to add enough acid to completely neutralize the first H and leave the second. That formula is too long to write; let's call it H2C. We add KOH to neutralize the first H to make KHC. Then we make the buffer out of the KHC and C^2-
mL x M = mL x M
500 x 0.0233 = mL x 0.202 which gives you about 11.65 (but you can do it more accurately and watch the significant figures). Then we make the buffer from that.
........HC^- + OH^- ==> C^2- + H2O
I.......11.65...0.......0........0
C........-x....-x.......x.........x
E.......11.65-x..0.......x........x

Plug all of this into the HH equation and solve for x.
2.75 = pK2 + log(base)/(acid)
base = x
acid = 11.65-x
x = mmols KOH. Convert to mL of the 0.202M stuff and ADD to the amount needed for the first equivalence point.
Then I would work backwards, plug in the numbers, and see if it really does produce a pH of 2.75.

1. 👍 0
2. 👎 0
posted by DrBob222
2. I ended up with KOH mL being about 1.5mL. The equiv. point would be around 57mL. I see the answer is to be around 20mL. I have no idea how this comes about.

1. 👍 0
2. 👎 0
posted by Micki

## Similar Questions

1. ### Analytical Chemistry

How many milliliters of 0.202 M KOH should be added to 500.0 mL of 0.0233 M tartaric acid (C4H6O6; FW 150.087) to adjust the pH to 2.75? Ka1 = 1.0 x 10-13 and Ka2 = 4.6 x10-5. I ended up with KOH mL being about 1.5mL. The equiv.

asked by Micki on March 6, 2013
2. ### Chemistry

A buffer is prepared by mixing 205 mL of .452 M HCl and .500 L of .400 M sodium acetate. Ka =1.80 x 10^-5 I got the pH is 4.79 but i can't figure out the grams of KOH the problem says how many grams of KOH must be added to .500 L

asked by Yumi on November 30, 2010
3. ### college chemistry

Consider the titration of a 50.0 mL sample of a 0.100 M solution of the triprotic weak acid citric acid (H3C6H5O7) with 0.100 M KOH. For citric acid, the three (3) acid dissociation constant values are ka1 = 7.40x10-3, ka2 =

asked by Aubree on April 21, 2010
4. ### Chemistry

A student titrates 0.025 M KOH into 50.00 ml of a solution of unknown weak acid HA. Equivalence point is reached when 30.00 ml of the KOH has been added. After 15.00ml of the KOH has been added the pH of the mixture is 3.92. A.

asked by Patrick on March 10, 2011
5. ### Chemistry

A student titrates 0.025 M KOH into 50.00 ml of a solution of unknown weak acid HA. Equivalence point is reached when 30.00 ml of the KOH has been added. After 15.00ml of the KOH has been added the pH of the mixture is 3.92. A.

asked by Patrick on March 10, 2011
6. ### chemistry

How many milliliters of water must be added to 153.2 mL of 2.73 M KOH to give a 1.68 M solution?

asked by walter on June 10, 2015
7. ### Chemistry

How many milliliters of a 0.20 M KOH are needed to completely neutralize 90.0 milliliters of 0.10 M HCl?

asked by Danny on May 5, 2010
8. ### chemistry

How many milliliters of 1.50 M \rm KOH solution are needed to provide 0.140 mol of \rm KOH

asked by betty on September 18, 2010
9. ### Chemistry

How many milliliters of 1.40 M KOH solution are needed to provide 0.120 mol of KOH? please explain how to solve

asked by Veronica on October 7, 2010
10. ### Chemistry

A buffer is prepared by mixing 205 mL of .452 M HCl and .500 L of .400 M sodium acetate. (Ka =1.80 x 10^-5) a) What is the pH? b) How many grams of KOH must be added to .500 L of the buffer to change the pH by .125 units?

asked by Missy on November 29, 2010

More Similar Questions